opepayflarpws

2022-06-22

I've been working on this question but just can't crack it. It is as follows.
Using $\frac{dy}{dx}=\frac{a\sqrt{{x}^{2}+{y}^{2}}+by}{bx}$ where y(1)=0
and by letting $v=\frac{y}{x}$ transform the above differential equation to an equation in v and x. Find the implicit solution of the resulting separable equation. Then use the relationship y=xv to obtain $y+\sqrt{{x}^{2}+{y}^{2}}={x}^{\frac{a}{b}+1}$
I've tried separate the upper terms into two fractions which makes the second term simply $\frac{by}{bx}=v$ but I am still left with the nasty square root with the y on the inside and I can't shift it easily to solve as a first order DE. Also I was thinking about solving it as a linear first order ODE but we have a squared term. Sorry about formatting as well, first time posting a question.

assumintdz

Just as you did
$\frac{dy}{dx}=\frac{a\sqrt{{x}^{2}+{y}^{2}}+by}{bx}$
Let
$y=xv\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime }=v+x{v}^{\prime }$
making the equation to be
$x{v}^{\prime }=\frac{a}{b}\sqrt{1+{v}^{2}}$
which is separable
$\int \frac{dx}{x}=\frac{b}{a}\int \frac{dv}{\sqrt{1+{v}^{2}}}$
making
$\mathrm{log}\left(x\right)+C=\frac{b}{a}{\mathrm{sinh}}^{-1}\left(v\right)$
So
$v=\mathrm{sinh}\left(\frac{a}{b}\mathrm{log}\left(x\right)+C\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y=x\phantom{\rule{thinmathspace}{0ex}}\mathrm{sinh}\left(\frac{a}{b}\mathrm{log}\left(x\right)+C\right)$

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